Given a prime number $p$ , establish the congruence:
$$(p-1)! \equiv (p-1) \pmod{1+2+3+\cdots+(p-1)}$$

I have proceeded like this:
$$\begin{align*}&(p-1)! \equiv (-1) \pmod{p} \quad \quad \quad \text{by Wilson's Theorem}\\
&(p-1)! \equiv 0 \pmod{\frac{p-1}{2}} \end{align*}$$
Then I know that I have to apply Chinese remainder theorem but I don't have a thorough understanding of it.So please give me an elaborate answer to this question with respect to Chinese remainder theorem.

One more way to think of this is that
$$
(p-1)!-(p-1)\equiv0\pmod{p}\tag{1}
$$
and
$$
(p-1)!-(p-1)\equiv0\quad\left(\text{mod }\frac{p-1}{2}\right)\tag{2}
$$
Since $\left(p,\frac{p-1}{2}\right)=1$, we have that
$$
(p-1)!-(p-1)\equiv0\quad\left(\text{mod }p\cdot\frac{p-1}{2}\right)\tag{3}
$$

First of all, to get the formalities out of the way, since $p$ and $\frac{p-1}{2}$ are relatively prime, there is a unique evaluation$\pmod {p\cdot\frac{p-1}{2}}$ which corresponds to the two remainders you have. Once you're there, we have
$$
(p-1) \equiv 0 \pmod{\frac{p-1}{2}}
$$
as well as
$$
(p-1) \equiv (-1) \pmod p
$$
Which means that $(p-1)$ and $(p-1)!$ fulfills the same relations, hence they must be congurent$\pmod {p\cdot\frac{p-1}{2}}$.